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Lecture 8: More on Operational Amplifiers (Op Amps) Input Impedance of Op Amps and Op Amps Using Negative Feedback: ● Consider a general feedback circuit as shown. ◆ Assume that the amplifier has input impedance Rin. ◆ We wish to find the input impedance R'in of the circuit including the effect of negative feedback. ◆ For the case of no feedback (B = 0) we have: Rin = Vin / Iin Iin = Vin / Rin ◆ If we include negative feedback (with B < 0) the input to the amplifier is: Amplifier Vin + BVout Vout Vin ◆ The input current is now: Vx Mixer A € I in = (Vin + BVout ) / R in ◆ We showed last week for a circuit with negative feedback: € Vout = AVin /(1− AB) ABVin B Vin + € 1− AB Iin = Rin feedback network Vin = Rin (1− AB) V = in Input impedance with negative feedback ′ Rin is much larger than the no feedback case. ′ = Rin (1− AB) Rin It is also possible to lower R'in with negative feedback. K.K. Gan € L8: More on Op Amps 1 ● Input impedance of non-inverting amplifier: ◆ The input voltage is directly connected to the op amp ☞ the input impedance is expected to be large. ◆ The typical input resistance of a 741 op amp is 2 MΩ (no feedback case). Vin ◆ Pick R1 = 1 kΩ and Rf = 50 kΩ + Vout ☞ amplifier gain G ~ Rf / R1 = 50 A (power connections not shown) B = 1/G = 0.02 R1 Rf The open loop gain (A) as a function of frequency for the 741 can be read off the spec sheets. ☞ Calculate the input impedance of the non-inverting amp vs. frequency: f (Hz) A Input Impedance R'in (Ω) 1 5 10 10 4 x109 103 103 4 x 107 + 6 6 Vout 10 1 2 x 10 (R of op amp) A ◆ Vin ● - (power connections not shown) R1 Input impedance of inverting amplifier: Rf ◆ Point A is at ground (a virtual ground) ☞ The input voltage does not actually "see" the op amp. ☞ The input impedance of this configuration is simply: ′ = Vin / I in = R1 R in ◆ If we use the same resistors as in the non-inverting amplifier (R1 = 1 kΩ and Rf = 50 kΩ) ☞ the input impedance of this amp is 1 kΩ, independent of frequency. € ■ Thus the inverting amp has a low input impedance. This is one of the practical drawbacks to this amplifier configuration. K.K. Gan L8: More on Op Amps 2 Output Impedance of Op Amps Using Negative Feedback: ● ● € The output impedance of a circuit is defined as: Rout = Vout / Iout We wish to see how the above expression is modified by negative feedback. ◆ Assume Vin is grounded. ◆ Assume we put a voltage V at the output of the amp. ☞ The feedback network puts BVout (B < 0) back to the input. ■ This voltage appears across the input impedance as Vd. V − AVd model of the op amp Iout = out Rout with negative feedback V − ABVout Vin = out Rout Vout Vd Vout (1− AB) AV R in d = R out Rout V = out ′ Rout ☞ € € K.K. Gan The new output impedance is greatly reduced: R ′ = out Rout 1− AB ❑ R'out → 0 as A → ∞. L8: More on Op Amps B 3 Op Amp Stability and Compensation ● A major reason for using negative feedback with op amps is to make the amp stable against oscillations. ◆ It is still possible to drive the amp into oscillation under certain conditions. ◆ From a previous lecture we derived the gain equation for amps with feedback: V A G = out = Vin 1− AB ■ Oscillations occur when AB → 1. ■ This can occur for positive feedback. ◆ In principle, the inverting input of the op amp adds a fraction (determined by the feedback network) € of the output to the input with a relative phase of 1800. ◆ However at high frequencies this phase shift decreases, eventually reaches zero ☞ the circuit can become unstable (i.e. oscillate). ◆ Since the op amp is made up of many resistors and capacitors ☞ we can model these phase shifts using RC networks. ◆ Recall for a low pass RC filter the gain and phase shift is given by: V 1 G = out = Vin 1+ (ωRC) 2 ◆ ◆ €◆ ◆ ◆ tanφ = −ωRC At frequencies above the break point (ωRC = 1) the gain falls off as 1/ω. This falls off is 20 dB for each factor of 10 (or 6 dB per octave) increase in the frequency. The phase shift rapidly converges to -π/2 or -900. The phase shift that we want to avoid is 1800. In terms of voltage gain a filter that has the gain falling off as 1/ω2 will produce a 1800 phase shift. K.K. Gan L8: More on Op Amps 4 ◆ The easiest way to visualize this problem is by imagining two low pass RC filters in series since the gains of filters are multiplicative (but additive in dBs). 80 dB A 20 dB/decade 6 dB/octave 60 dB 40 dB Open loop gain (A) vs. frequency curve for a "typical" op amp -1/B 20 dB 40 dB/decade 12 dB/octave 10 ◆ ◆ ◆ 1 10 3 10 5 f (Hz) For 20 and 40 dB lines the frequency (x axis) at which the lines hit the gain curve is where A = -1/B. ■ If the phase shift at this frequency is 1800 oscillations will occur. For the 40 dB line ■ no oscillations can occur ■ the gain rolloff is only 20 dB/decade. ☞ the phase shift ≤ 900 For the 20 dB line ■ oscillations can occur ■ the gain rolloff is 40 dB/decade ☞ a 1800 phase shift is possible K.K. Gan L8: More on Op Amps 5 ● Compensation: ◆ To make an op amp stable against oscillation ■ make insure the open loop gain (A) falls off no faster than 20 dB/decade ☞ not possible to have a 1800 phase shift. ◆ Some op amps (e.g. µA741) are internally compensated (with capacitors) to insure that the gain roll-off is 20 dB or smaller all the way down to voltage gains of unity. ◆ A second type of op amp is called uncompensated ■ user adds compensating capacitors external to the op amp for stability against oscillation. ■ advantage: achieve higher gain by a suitable choice of capacitors ■ disadvantage: the circuit will oscillate if the wrong capacitor(s) was chosen! K.K. Gan L8: More on Op Amps 6